I am publishing the first part of the second chapter of lectures on the theory of automatic control.
This article discusses:


2.1. Obtaining equations of system dynamics. Static characteristic. The dynamics equation of self-propelled guns (ATS) in deviations
2.2. Linearization of the equations of dynamics ACS (ATS)
2.3. The classic way to solve dynamic equations


Lectures on the course "Management of Technical Systems", read by Oleg Kozlov at the Department of Nuclear Reactors and Power Plants, Faculty of Power Engineering, MSTU. N.E. Bauman. For which many thanks to him.


These lectures are only being prepared for publication in the form of a book, and since there are TAU specialists, students and just interested in the subject, any criticism is welcome.


First part: "Introduction to the theory of automatic control. Basic concepts of the theory of control of technical systems ”


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2.1. Obtaining equations of system dynamics. Static characteristic. The equation of dynamics of self-propelled guns (ATS) in deviations


When compiling equations describing non-stationary processes in an automatic control system (ATS) and which we will call dynamics equations in the future, the system is “divided” into separate elements (links), for each of which there are no problems in writing the corresponding dynamics equation.


In fig. 2.1.1 a schematic representation of the ACS (link) in the input-output variables is presented, where x (t) (or u (t) ) is the input action, and y (t) - output effect, respectively. Often, the input action will be called the manager , and the output effect will be called the controlled variable (variable).


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Fig. 2.1.1 - Schematic representation of self-propelled guns (links)

When compiling the equations of dynamics, fundamental conservation laws are used from the sections “Mechanics”, “Physics”, “Chemistry”, etc.


For example, when describing the movement of a node of some power drive mechanism, conservation laws are used: moment, energy, momentum, etc.... In thermophysical (thermohydraulic) systems, fundamental conservation laws are used: mass (continuity equation), momentum (equation of motion), energy (energy equation), etc.


The conservation equations in the general case contain constant and non-stationary members, and when discarding non-stationary members, they get the so-called static equations , which correspond to the equations equilibrium state self-propelled guns (link). Subtracting from the complete equations of conservation of stationary equations non-stationary equations of self-propelled guns in deviations (from the hospital).


$ \ left [\ begin {gathered} y (t)=y_0 + \ Delta y (t) \\ u (t)=u_0 + \ Delta u (t) \\ \ end {gathered} \ right. $


where: $ y_0, u_0 $- stationary values ​​of input and output actions;
$ \ Delta y, \ Delta u $- deviations from the station, respectively.

As an example, consider the "technology" of obtaining the equations of dynamics for a mechanical damper, a schematic representation of which is shown in Fig. 2.1.2.


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Fig. 2.1.2 - Mechanical damper

According to Newton’s 2nd law, the acceleration of the body is proportional to the sum of the forces acting on the body:

$ m \ cdot \ frac {d ^ 2 y (t)} {dt}=\ sum F_j \ \ \ \ mathbf {(2.1.1)} $


where, m - body mass, F j - all the forces acting on the body (damper piston)


Substituting in equation (2.1.1) all the forces according to Fig. 2.2, we have:


$ m \ cdot \ frac {d ^ 2y (t)} { dt}=m \ cdot g + u (t) - k \ cdot y (t) - c \ cdot \ frac {dy (t)} {dt} \ \ \ \ mathbf {(2.1.2)} $

where $ m \ cdot g $ - gravity; $ k \ cdot y (t) $- spring resistance force, $ c \ cdot \ frac {dy (t)} { dt} $ - viscous friction force (proportional to the piston speed)


Dimensions of forces and coefficients included in equation (2.1.2):

$ m \ cdot g \ Rightarrow [\ frac {kg \ cdot m } {s ^ 2}]; c \ Rightarrow [\ frac {kg} {s}]; k \ Rightarrow [\ frac {kg} {s ^ 2}] $

Assuming that at t ≤ 0, the damper piston was in equilibrium, that is,

$$ display $$ \ left \ {\ begin {eqnarray} t & amp; \ le 0 \\ u (t) & amp;=u_0 \ \ y (t) & amp;=y_0 \\ \ end {eqnarray} \ right. $$ display $$

move on to the deviations from the stationary state:
Suppose that at t > 0 $ \ left [\ begin {gathered} y (t)=y_0 + \ Delta y (t); u (t)=u_0 + \ Delta u (t); \\ y '(t)=[\ Delta y (t)]; y' '(t)=[\ Delta y (t)] '\\ \ end {gathered} \ right. $ . Then, substituting these relations into equation (2.1.2), we obtain:

$ m \ cdot \ frac {d ^ 2 \ Delta y (t )} {dt ^ 2}=m \ cdot g - k \ cdot y_0- k \ cdot \ Delta y (t) - c \ frac {d \ Delta y (t)} {dt} \ \ \ \ mathbf {( 2.1.3)} $


if $ t \ le 0 $, then the equation takes the form:

$ 0=m \ cdot g + u_0 - k \ cdot y_0 \ \ \ \ mathbf {(2.1.4)} $

or

$ y_0=\ frac {1} {k} \ cdot [u_0 + m \ cdot g] \ \ \ \ mathbf {(2.1.5)} $


Relation (2.1.4) is the equation of the link (damper) in the equilibrium (stationary) state, and relation (2.1.5) is the static characteristic of the link - damper (see Figure 2.1.3).


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Fig. 2.1.3 - Static characteristic of a mechanical damper

Subtracting equation (2.1.4), we obtain the equation of dynamics of the damper in the deviations:

$ m \ cdot g \ frac {d ^ 2 \ Delta y ( t)} {dt ^ 2}=\ Delta u (t) - k \ cdot \ Delta y (t) - c \ cdot \ frac {d \ Delta y (y)} {dt}, $


then, dividing by k, we have:

$ T_2 ^ 2 \ cdot \ frac {d ^ 2 \ Delta y (t)} {dt ^ 2} + T_1 \ cdot \ frac {d \ Delta y (t)} {dt} + \ Delta y (t)=k_1 \ cdot \ Delta u (t) \ \ \ mathbf { (2.1.6)} $


where:

$ T_2 ^ 2=\ frac {m} {k}; T_1=\ frac {c} {k}; k_1=\ frac {1} {k}. $


Equation (2.1.6) is an equation of dynamics in canonical form, i.e. the coefficient at Δy (t) is 1.0!


It is “easy” to see that the coefficients in front of the terms containing derivatives make sense (and dimension!) of time constants . In fact:


$ T_1=\ frac {c} {k} \ Rightarrow [\ frac {kg \ cdot c ^ 2} {c \ cdot kg}]=[c] $


$ T_2 ^ 2=\ frac {m} {k} \ Rightarrow [\ frac {kg \ cdot c ^ 2} {kg}]=[c ^ 2] $


Thus, we get that:
- the coefficient before the first derivative has the dimension [c] ie the meaning of a certain time constant;
- coefficient before the second derivative: [ $ c ^ 2 $];
- coefficient on the right side ( $ k_1 $): [ $ \ frac {c ^ 2} {kg} $].
Then equation (2.1.6) can be written in the operator form:


$ (T_2 ^ 2 \ cdot p ^ 2 + T_1 \ cdot p + 1) \ Delta y (t)=k_1 \ Delta u (t) $ , which is equivalent to

$ L (p) \ Delta y (t)=N (p ) \ Delta u (t) \ \ \ \ mathbf {(2.1.6.a)} $

where:  $ p=\ frac {d} {dt} $ - differentiation operator;
$ L (p)=T_2 ^ 2 \ cdot p + T \ cdot p + 1 $ -linear differential operator; $ L (p) $
$ N (p) $- linear differential operator degenerated into a constant equal to $ k_1 $.

Analysis of equation (2.1.6.a) shows that such an equation has dimensional variables, and all coefficients of the equation are dimensional. This is not always convenient. In addition, if a real ATS (ACS) consists of many links, the output actions of which are various physical variables (speed, temperature, neutron flux, heat flux, etc.), then the coefficient values ​​can vary by a large number of orders, which puts serious mathematical problems in the numerical solution of the equations of dynamics on a computer (since numbers in a computer are always represented with some accuracy). One of the best ways to avoid numerical difficulties is the principle of normalization, i.e. the transition to dimensionless deviations, which are obtained by normalizing the deviations to the stationary value of the corresponding variable.


Введем новые нормированные (безразмерные) переменные:

$\left[ \begin{gathered} \widetilde y(t)=\frac {y(t)-y_0}{y_0}=\frac {\Delta y(t)}{y_0}\\ \widetilde u(t)=\frac {u(t)-u_0}{u_0} =\frac {\Delta u(t)}{u_0}\\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered} y(t) =y_0 \cdot [1+ \widetilde y(t)]\\ u(t)=u_0 \cdot [1+ \widetilde u(t)]\\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered} y'(t) =y_0 \cdot \widetilde y(t)\\ y''(t)=y_0 \cdot \widetilde y(t)''\\ \end{gathered} \right. $


Подставляя эти соотношения в уравнение (2.1.2), имеем:

$m \cdot y_0 \cdot \frac{d^2 \widetilde y(t)}{dt^2}=m \cdot g +u_0 \cdot[1+\widetilde u(t)] -k \cdot y_0 \cdot[1+\widetilde y(t)] - c \cdot y_0 \cdot \frac{d \widetilde y(t)}{dt}; или$

$m \cdot y_0 \cdot \frac{d^2 \widetilde y(t)}{dt^2}=\underline {m \cdot g} + \underline {u_0} +u_0 \cdot \widetilde u(t) - \underline {k \cdot y_0} - k \cdot y_0 \cdot \widetilde y(t) - c \cdot y_0 \cdot \frac{d \widetilde y(t)}{dt}.$


Поддчеркнутые члены выражения в сумме дают 0 (см. 2.1.4) Перенося в левую часть члены, содержащие $\widetilde y(t)$, и, разделив на $k \cdot y_0$, получаем:

$\frac{m \cdot y_0}{k \cdot y_0} \cdot \frac{d^2 \widetilde y(t)}{dt^2} + \frac {c \cdot y_0}{k \cdot y_0} \cdot \frac{d \widetilde y(t)}{dt} +\widetilde y(t)= \frac{u_0}{k \cdot y_0} \cdot \widetilde u(t) \Rightarrow $

$(T_2^2 \cdot p + T \cdot p + 1) \cdot \widetilde y(t)=k_x \cdot \widetilde u(t) \ \ \ \mathbf{(2.1.7)}$


где: $T_2^2= \frac{m}{k}; T_1= \frac{c}{k}; k_x=\frac{u_0}{k \cdot y_0} $— коэффициент усиления, причем безразмерный.

Проверим размерность коэффициента $k_x$

$\left[\frac{кг \cdot м}{с^2} \cdot \frac{c^2}{кг} \cdot \frac{1}{м}\right] =[1].$


Использованный выше «технический» прием позволяет перейти к безразмерным переменным, а также привести вид коэффициентов в уравнении динамики к легко интерпретируемому виду, т.е. к постоянным времени (в соответствующей степени) или к безразмерным коэффициентам усиления.


На рис. 2.1.4 представлены статические характеристики для механического демпфера:


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Рис. 2.1.4 – Статические характеристики механического демпфера

Процедура нормировки отклонений позволяет привести уравнения динамики к виду:

$L(p) \cdot \widetilde y(t)=N(p) \cdot \widetilde u(t) \ \ \ \mathbf{(2.1.8)}$

где $L(p), N(p) - $дифференциальные операторы.

If the differential operators $ L (p), N (p) - $ is linear , and the static characteristic of the ACS (link) is also linear, then expression (2.1.8) corresponds to the linear ordinary differential equation ( ODE).


And if $ L (p), N (p) - $- nonlinear differential operators, or $ \ widetilde y_ {stat} \ neq k \ cdot \ widetilde u_ {stat} $ , then the equation of dynamics is non-linear. Non-linear actions are understood as all mathematical operations, except addition (+) and subtraction (-).


An example of creating a damper model can be seen here: "Technology for obtaining equations of dynamics of TAU"



2.2. Linearization of the equations of dynamics of ACS (ATS)


Almost all real automatic control systems (ACS) are non-linear, and non-linearity of ACS can be determined by various reasons:


  1. The non-linearity of the static characteristic.
  2. The nonlinearity of the dynamic terms in the equations of dynamics.
  3. The presence in the self-propelled guns of fundamentally non-linear links.

If in a closed ACS (ATS) there are no fundamentally non-linear links , then in most cases the equations of dynamics of the links included in the system can be linearized. Linearization is based on the fact that in the process of regulation (i.e., self-propelled guns with feedback), all controlled quantities deviate little from their program values ​​(otherwise, the control or control system would not fulfill its task).


For example, if we consider the power control of a nuclear power reactor, then the main task of the ATS is to maintain power at a given (nominal) power level. Existing perturbations (internal and external) are “worked out” by the ATS and therefore the parameters of the nuclear reactor are slightly different from the stationary ones. In fig. 2.2.1 shows the time dependence of the power of a nuclear reactor, where the normalized power deviations are ΔN/N 0 & lt; & lt; 1, and therefore, the equations of dynamics of a nuclear reactor, in principle, can be linearized.


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Fig. 2.2.1 - An example of a change in reactor power

Consider some link (or CAP as a whole), a description of the dynamics of which can be represented in the input-output variables:


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Fig. 2.2.2 - ATS Link

Suppose that the dynamics of this link is described by an ordinary differential equation of the nth order:

$ L (p) \ cdot y (t)=N (p ) \ cdot u (t) \ \ \ \ mathbf {(2.2.1)} $


Let's transfer $ N (p) u (t) $to the left side of the equation and write the equation as%

$ F (y, y ', y' ',... y ^ {n}, u, u ', u' ',... u ^ m, t)=0 \ \ \ \ mathbf {(2.2.2)} $


where $ F $-– the function of the controlled variable and its derivatives, as well as the control (input) action and its derivatives, and F is usually a nonlinear function.


We assume that at t ≤ 0 the ACS (link) was in equilibrium (in a stationary state). Then the equation (2.2.2) degenerates into the equation of static characteristic:

$ y '= y' '=... y ^ n=u '= u' '=... u ^ m=0 \ Rightarrow F (y_0, u_0)=0 \ \ \ \ mathbf {(2.2.3)} $


We expand the left side of equation (2.2.2) in a Taylor series in a small neighborhood of the equilibrium state point $ (y_0, u_0) (y0, u0) $.


Recall that Taylor expansion is interpreted as follows: if $ y=f (x) $ , then the" simple "expansion of the function in a Taylor series in the neighborhood of the  $ x=x_0 $ will look like this:


$ \ begin {eqnarray} y (x)=f (x)=f (x_0) + \ frac {1} {1!} \ cdot \ left (\ frac {df} {dx} \ right) _ {x=x_0} \ cdot (x-x_0) + \ frac {1} {2!} \ Cdot \ left (\ frac {d ^ 2f} {dx ^ 2} \ right) _ {x=x_0} \ cdot (x-x_0) + \\... + \ frac {1} { n!} \ cdot \ left (\ frac {d ^ nf} {dx ^ n} \ right) _ {x=x_0} \ cdot (x-x_0) \ end {eqnarray} $


In view of the above, the decomposition takes the form:


$ \ begin {eqnarray} F (y, y ', y' ',... y ^ {n}, u, u',... u ^ m, t)=F (y_0, u_0) + \ frac {1} {1!} \ cdot \ left (\ frac { dF} {dy} \ right) _ {y=y_0; u=u_0} \ cdot (y-y_0) + \\ + \ frac {1} {2!} \ cdot \ left (\ frac {d ^ 2F} {dy ^ 2} \ right) _ {y=y_0; u=u_0} \ cdot (y-y_0) ^ 2 + \ frac {1} {3!} \ cdot \ left (\ frac {d ^ 3F} { dy ^ 3} \ right) _ {y=y_0; u=u_0} \ cdot (y-y_0) ^ 3 +.. \\ + \ frac {1} {k!} \ cdot \ left (\ frac {d ^ kF} {dy ^ k} \ right) _ {y=y_0; u=u_0} \ cdot (y-y_0) ^ k + \ frac {1} {1!} \ cdot \ left (\ frac {dF} { dy '} \ right) _ {0} \ cdot (y') + \ frac {1} {2!} \ cdot \ left (\ frac {d ^ 2F} {d (y ') ^ 2} \ right) _ {0} \ cdot (y ') ^ 2 + \\.. + \ frac {1} {1!} \ Cdot \ left (\ frac {dF} {dy ^ {(n)}} \ right) _ {0} \ cdot (y ^ {(n)}) + \ frac {1} {2!} \ Cdot \ left (\ frac {d ^ 2F} {d (y ^ {n}) ^ 2} \ right ) _ {0} \ cdot (y ^ {(n)}) ^ 2 +.. \\.. + \ frac {1} {1!} \ Cdot \ left (\ frac {dF} {du} \ right ) _ {0} \ cdot (u-u_0) + \ frac {1} {2!} \ Cdot \ left (\ frac {d ^ 2F} {du ^ 2} \ right) _ {0} \ cdot (u -u_0) ^ 2 +.. \\.. + \ frac {1} {k!} \ cdot \ left (\ frac {d ^ k F} {d (u ^ {m}) ^ k} \ right) _ {0} \ cdot (u ^ {(m)}) ^ k +.. \ end {eqnarray} $


Assuming that the deviations of the output and input effects are insignificant, (i.e.: $ \ frac {\ Delta y (t)} {y_0} & lt; & lt; 1; \ frac {\ Delta u (t)} {u_0} & lt; & lt; 1; $), we leave in the expansion only the terms of the first order of smallness (linear). Because $ [y (t) -y_0] \ equiv \ Delta y (t) ; y '(t)=[y_0 + \ Delta y (t)]'=(\ Delta y (t)) '$ , we get:


$ F (y, y ', y' '.. y ^ {(n)}, u, u ', u' '.. u ^ {m}) \ simeq \ left (\ frac {\ partial F} {\ partial y} \ right) _0 \ cdot \ Delta y + \ left (\ frac {\ partial F} {\ partial y '} \ right) _0 \ cdot (\ Delta y)' + \ left (\ frac {\ partial F} {\ partial y ''} \ right) _0 \ cdot (\ Delta y) '' +.. \\.. + \ left (\ frac {\ partial F} {\ partial u} \ right) _0 \ cdot \ Delta u + \ left (\ frac {\ partial F} { \ partial u '} \ right) _0 \ cdot (\ Delta u)' + \ left (\ frac {\ partial F} {\ partial u ''} \ right) _0 \ cdot (\ Delta u) '' +.. + \ left (\ frac {\ partial F} {\ partial u ^ {m}} \ right) _0 \ cdot (\ Delta u) ^ {(m)} \ \ \ \ mathbf {(2.2.4)} $


Substituting relation (2.2.4) into equation (2.2.2), и перенося множители при у и u в разные части получаем уравнения:

$a_n^0 \cdot \frac{d^{(n)}}{dt^n} \Delta y +a_{n-1}^0 \cdot \frac{d^{(n-1)}}{dt^{n-1}} \Delta y +..+a_{1}^0 \cdot \frac{d}{dt} \Delta y +a_{0}^0\cdot\Delta y=\\ =b_m^0 \cdot \frac{d^{(m)}}{dt^m} \Delta u +b_{m-1}^0 \cdot \frac{d^{(m-1)}}{dt^{m-1}} \Delta u +..+b_{1}^0 \cdot \frac{d}{dt} \Delta u +b_{0}^0\cdot\Delta u \ \ \ \mathbf{(2.2.5)}$

где:

$a_0^0= \left( \frac{\partial F}{\partial y} \right)_{y =y_0;u =u_0};a_1^0= \left( \frac{\partial F}{\partial y'} \right)_{y =y_0;u =u_0};...a_n^0= \left( \frac{\partial F}{\partial y^{(n)}} \right)_{y =y_0;u =u_0} ; \\ b_0^0= \left( \frac{\partial F}{\partial u} \right)_{y =y_0;u =u_0};b_1^0= \left( \frac{\partial F}{\partial u'} \right)_{y =y_0;u =u_0};...b_m^0= \left( \frac{\partial F}{\partial u^{(m)}} \right)_{y =y_0;u =u_0}.$


Коэффициенты $a_i^0, b_j^0$— постоянные коэффициенты, поэтому уравнения 2.2.5 — линейное дифференциальное с постоянными коэффициентами.


В дальнейшем нами будет часто использоваться операторная форма записи уравнений динамики:

$L(p)\cdot\Delta y(t)=N(p) \cdot \Delta u(t)\ \ \ \mathbf{(2.2.6)}$


где $p=\frac{\partial }{\partial t}$– оператор дифференцирования;
$L(p)$— линейный дифференциальный оператор степени n;
$N(p)$— линейный дифференциальный оператор степени m, причем обычно порядок оператора $L(p)$выше порядка оператора $N(p)$: $ n ≥ m.$

Уравнения (2.2.5) и (2.2.6) — уравнения динамики системы (звена) в отклонениях.


Если исходное уравнение (2.2.1) — дифференциальное уравнение в физических переменных (температура, скорость, поток и т.д.), то размерность коэффициентов $a_i^0, b_j^0$может быть произвольной (любой).


Переход к нормализованным отклонениям позволяет “упорядочить” размерность коэффициентов. В самом деле, разделив уравнение (2.2.5) на $(y_0, u_0)$и выполнив некоторые преобразования, получаем:


$a_n^* \cdot \tilde{y}^{(n)} +a_{(n-1)}^* \cdot \tilde{y}^{(n-1)} +..+a_1^* \cdot \tilde{y}'+a_0^* \cdot \tilde{y}=\\ =b_m^* \cdot \tilde{u}^{(m)} +b_{(m-1)}^* \cdot \tilde{u}^{(m-1)} +..+b_1^* \cdot \tilde{u}'+b_0* \cdot \tilde{u} \ \ \ \mathbf{(2.2.7)}$


Приведение уравнения динамики САУ (звена) к нормализованному виду позволяет “унифицировать” размерность коэффициентов уравнений: ==>


$$display$$[a_0^*]=[1] ;\ \ [a_1^*]= [c];\ \ [a_2^*]= [c^2];\ \ [a_3^*]= [c^3];.[a_n^*]= [c^n]\\ [b_0^*]=[1] ;\ \ [b_1^*]= [c];\ \ [b_2^*]= [c^2];\ \ [b_3^*]= [b^3];.[b_m^*]= [c^m]$$display$$


Если вынести в правой части (2.2.7) коэффициент $b_0^*$за общую скобку и разделить все уравнение на $a_0^*$, то уравнение принимает вид:


$a_n \cdot \tilde{y}^{(n)} +a_{(n-1)}\cdot \tilde{y}^{(n-1)} +..+a_1\cdot \tilde{y}'+\tilde{y}=\\ =k \cdot [b_m \cdot \tilde{u}^{(m)} +b_{(m-1)} \cdot \tilde{u}^{(m-1)} +..+b_1 \cdot \tilde{u}'+ \tilde{u}] \ \ \ \mathbf{(2.2.8)}$


где:

$a_n=\frac{a_n^*}{a_0^*};\ a_{n-1}=\frac{a_{n-1}^*}{a_0^*}; \...a_{1}=\frac{a_{1}^*}{a_0^*}; \ k=\frac{b_0^*}{a_0^*} \\ b_n=\frac{b_n^*}{b_0^*};\ b_{n-1}=\frac{b_{n-1}^*}{b_0^*}; \...b_{1}=\frac{b_{1}^*}{b_0^*}; $

или в операторном виде:

$(a_n \cdot p^{(n)} +a_{(n-1)}\cdot p^{(n-1)} +..+a_1\cdot p'+1) \cdot \tilde{y}=\\ =k \cdot (b_m \cdot p^{(m)} +b_{(m-1)} \cdot p^{(m-1)} +..+b_1 \cdot \tilde{u}'+ 1)\cdot \tilde{u}\\ L(p)\cdot \tilde{y} =k \cdot N(p) \cdot \tilde{u} \ \ \ \mathbf{(2.2.9)}$


Линеаризация уравнений динамики и нормализация переменных позволяют привести уравнения динамики САУ (звена) к виду, наиболее удобному для использования классических методов анализа, т.е. к нулевым начальным условиям .


$$display$$Если \ t ≤ 0 \Rightarrow \left[ \begin{gathered} \tilde {y}(t)=\tilde {y}(0) =0;\\ \tilde u(t)=\tilde u(0)=0.\\ \end{gathered} \right.$$display$$


Пример


Выполнить линеаризацию уравнения динамики некоторой «абстрактной» САР в окрестности состояния (x 0 , y 0 ) , если полное уравнение динамики имеет вид:


$a_3^0 \cdot y'''(t) +a_2^0 \cdot y''(t)+a_1^0 \cdot y'(t) \cdot[y(t)-d]+ a_2^0 \cdot y^2(t)=b_1^0 \cdot x'(t) +b_0^0 \cdot x(t); $


Нелинейность полного уравнения динамики проявляется в следующем:


• во-первых, в нелинейности статической характеристики:


$a_0^0 \cdot y^2(0)=b_0^0 \cdot x(0); \Rightarrow y_0=\sqrt{\frac{b_0^0}{a_0^0} \cdot x_0}=\sqrt {k_0 \cdot x_0}$


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Рис. 2.2.3 – Линеаризации статической характеристики

• во-вторых, слагаемое в левой части $a_1^0 \cdot y'(t) \cdot[y(t)-d]$— чисто нелинейное, так как действие умножения является нелинейным.


Выполним процесс линеаризации исходного уравнения, динамики без разложения я ряд Тейлора, основываясь на том, что в окрестности состояния (x 0 , y 0 ) нормированные отклонения управляющего воздействия и регулируемой величины намного меньше 1.


Преобразования выполним в следующей последовательности:


  1. Перейдем к безразмерным переменным (нормализованным);
  2. Выполним линеаризацию, отбросив нелинейные члены 2-го и выше порядков малости.

Перейдем к новым безразмерным переменным:


$\tilde{y}(t)=\frac{y(t) - y_0}{y_0};\ \Rightarrow y(t)=y_0 \cdot [1+ \tilde{y}(t)]; \\ \tilde{x}(t)=\frac{x(t) - x_0}{x_0};\ \Rightarrow x(t)=x_0 \cdot [1+ \tilde{x}(t)].$


Заметим, что:
$x(t)=x_0+ \Delta x(t)=x_0+ x_0 \cdot \tilde{x}(t) \ и \ y(t)=y_0+ \Delta y(t)=y_0+ y_0 \cdot \tilde{y}(t)$.


Подставляя значения x(t) и y(t) в исходное уравнение:


$ a_3^0 \cdot y_0 \cdot \tilde y'''(t) +a_2^0 \cdot y_0 \cdot \tilde y''(t)+a_1^0 \cdot y_0 \cdot \tilde y'(t) \cdot[y_0+y_0 \cdot \tilde y(t) -d]+a_0^0 \cdot y_0^2 \cdot[1+ \tilde y(t)]^2=\\=b_1^0 \cdot x_0 \cdot \tilde x'(t) + b_0^0 \cdot х_0 \cdot[1+\tilde x(t)]; \ \Rightarrow \ раскрыв \ скобки \Rightarrow \\ a_3^0 \cdot y_0 \cdot \tilde y'''(t)+a_2^0 \cdot y_0 \cdot \tilde y''(t)+a_1^0 \cdot y_0^2 \cdot \tilde y'(t) + a_1^0 \cdot y_0^2 \cdot \tilde y'(t) \cdot \tilde y(t)-a_1^0 \cdot y_0 \cdot \tilde y'(t) \cdot d + \\ +a_0^0 \cdot y_0^2 +2 \cdot a_0^0 \cdot y_0^2 \cdot\tilde y(t) +a_0^0 \cdot y_0^2 \cdot\tilde y(t)^2=b_1^0 \cdot x_0 \cdot \tilde x'(t) + b_0^0 \cdot х_0+ b_0^0 \cdot х_0 \cdot \tilde x(t)];$


Удаляем полученного уравнения уравнения стационара: $a_0^0 \cdot y_0^2= b_0^0 \cdot х_0$, а так же пренебрегая слагаемыми второго прядка малости: $\ a_1^0 \cdot y_0^2 \cdot \tilde y'(t) \cdot \tilde y(t)$, получаем следующее уравнение:


$ a_3^0 \cdot y_0 \cdot \tilde y'''(t)+a_2^0 \cdot y_0 \cdot \tilde y''(t)+(a_1^0 \cdot y_0^2 -a_1^0 \cdot y_0 \cdot d) \cdot \tilde y'(t)+2 \cdot a_0^0 \cdot y_0^2 \cdot\tilde y(t) =\\=b_1^0 \cdot x_0 \cdot \tilde x'(t) + b_0^0 \cdot х_0 \cdot \tilde x(t);$


Вводим новые обозначения:

$a_3^*=a_3^0 \cdot y_0 ; \ \ a_2^*=a_2^0 \cdot y_0; \ \ a_1^*=a_1^0 \cdot y_0^2 -a_1^0 \cdot y_0 \cdot d; \ \ a_0^* =2 \cdot a_0^0 \cdot y_0^2; \\ b_1^*=b_1^0 \cdot x_0; \ \ b_0^*=b_0^0 \cdot х_0 \ $


Получаем уравнения в «почти» классическом виде:


$a_3^* \cdot \tilde y'''(t)+a_2^* \cdot \tilde y''(t)+a_1^* \cdot \tilde y'(t)+ a_0^* \cdot\tilde y(t)=b_1^* \cdot \tilde x'(t) + b_0^* \cdot \tilde x(t);$


Если в правой части вынести за общую скобку $b_0^*$и разделить все уравнение на $a_0^*$, то уравнение (линеаризованное) принимает вид:


$a_3 \cdot \tilde y'''(t)+a_2 \cdot \tilde y''(t)+a_1 \cdot \tilde y'(t)+ \tilde y(t)=k \cdot[ b_1 \cdot \tilde x'(t) + \tilde x(t)]$


где:

$a_3=\frac{a_3^*}{a_0^*}; \ \ a_2=\frac{a_2^*}{a_0^*}; \ \ a_1=\frac{a_2^*}{a_0^*}; \ \ k=\frac{b_o^*}{a_0^*}; \ \ b_1=\frac{b_1^*}{b_0^*}; $


Процедура нормализации позволяет более просто линеаризовать уравнение динамики, так как не требуется выполнять разложение в ряд Тейлора (хотя это и не сложно).


2.3. Классический способ решения уравнений динамики


Классический метод решения уравнений динамики САУ (САР) применим только для линейных или линеаризованных систем.


Consider some self-propelled guns (link), the dynamics of which are described by a linear differential equation of the form:


$ L (p) \ cdot y (t)=N (p ) * x (t), \ \ \ \ mathbf {(2.3.1)} \\ where: L (p)=a_n \ cdot p ^ n +.. + a_1 \ cdot p + a_0 \\ N (p)=b_m \ cdot p ^ m +.. + b_1 \ cdot p + b_0 $


Turning to the full symbolism, we have: $ \ Rightarrow $


$ a_n \ cdot y ^ {(n)} + a_ {n -1} \ cdot y ^ {(n-1)} +.. + a_1 \ cdot y '+ a_ {0}=b_m \ cdot x ^ {(m)} + b_ {n-1} \ cdot x ^ {(n-1)} +.. + b_1 \ cdot y '+ b_ {0} \ \ \ mathbf {(2.3.2)}; $


Expression (2.3.2) is an ordinary differential equation (ODE), more precisely, an inhomogeneous ODE, since the right-hand side is ≠ 0.


The input action x (t), the coefficients of the equation, and the initial conditions are known (i.e., the values ​​of the variables and derivatives at t=0).


It is required to find y (t) under known initial conditions.


It is known that


$ y (t)=y_ {total} (t) + y_ {frequent} (t), \ \ \ \ mathbf {(2.3.3)} $


where: $ y_ {total} (t) $- solution of the homogeneous differential equation $ L (p) y ( t)=0; $ y_ {part} (t) $ inline $ is a private solution. $ inline $


We will call the solution of the homogeneous differential equation $ y_ {common}=y_ {own} $ , by its own solution, since its solution does not depend on the input impact, but is completely determined by own dynamic properties of self-propelled guns (link).


The second component of the solution (2.3.3) will be called $ y_ {frequent. }=y_ {dec.} $ , forced, because this part of the solution is determined by the external influence  $ x (t) $ , so the self-propelled guns (ATS or link) are“ forced to work out ”this impact:


$ y (t)=y_ {property} (t) + y_ {dec.} (t), \ \ \ \ mathbf {(2.3.4)} $


Recall the steps of the solution:


1) If there is an equation of the form $ L (p) \ cdot y ( t)=N (p) * x (t) $ , then first we solve the homogeneous differential equation:


$ a_n \ cdot y ^ {(n)} + a_ {n -1} \ cdot y ^ {(n-1)} +.. + a_1 \ cdot y '+ a_ {0}=0 $


2) We write the characteristic equation:


$ L (\ lambda)=0; \ \ Rightarrow \ a_n \ cdot \ lambda ^ {n} + a_ {n-1} \ cdot \ lambda ^ {n-1} +.. + a_1 \ cdot \ lambda + a_ {0}=0 \ \ \ \ mathbf {(2.3.5 )} $


3) Solving the equation (2.3.5), which is a typical power equation, in some way (including using standard routines on a computer) we find the roots of the characteristic equation  $ \ lambda_i $
4) Then your own decision is written in the form:


$ y_ {property} (t)=\ sum_ {j=1} ^ n C_j \ cdot e ^ {\ lambda_j \ cdot t}, \ \ \ \ mathbf {(2.3.6)} $


if among $ \ lambda_i $no duplicate roots (the multiplicity of roots is 1).


If equation (2.3.5) has two coinciding roots, then its own solution has the form:


$ y_ {property} (t)=\ sum_ {j=1} ^ {n-2} C_j \ cdot e ^ {\ lambda_ {j} \ cdot t} + C_ {n-1} \ cdot e ^ {\ lambda_ {n-1} \ cdot t} \ cdot [1 + C_n \ cdot t]. \ \ \ \ Mathbf {(2.3.7)} $


If equation (2.3.5) has k coinciding roots (the multiplicity of roots is k), then the own solution has the form:


$ y_ {property} (t)=\ sum_ {j=1} ^ {nk} C_j \ cdot e ^ {\ lambda_ {j} \ cdot t} + C_ {n + 1-k} \ cdot e ^ {\ lambda_ {n + 1-k} \ cdot t} \ cdot [1 + C_ {n + 2-k} \ cdot t + C_ {n + 3-k} \ cdot t ^ 2 +.. \\.. + C_ {n} \ cdot t ^ {k-1}]. \ \ \ \ mathbf {(2.3.8)} $


5) The forced part of the solution can be found in various ways, but the following methods are most common:
a) By the appearance of the right side.
b) The method of variation of constants.
c) Other methods...


If the form of the right side of the differential equation is a relatively simple function of time, then method a) is preferred: selection of a solution. $ y_ {dec.} (t)=f_ {dec} (t) $ .


6) Summing up the obtained components (proper and forced), we have: $ \ Rightarrow $


$ y_ {full} (t)=\ sum_ {j=1} ^ n C_j \ cdot e ^ {\ lambda_j \ cdot t} + f_ {out} (t). $


7) Using the initial conditions (t=0), we find the values ​​of the integration constants  $ C_j $ . $ \ Rightarrow $Usually we get a system of algebraic equations. $ \ Rightarrow $Solving the system, we find the values integration constant $ C_j $


Example


Find the analytical expression of the transient at the output of the link, if


$ \ left \ {\ begin {gathered} 2 \ cdot y ' '(t) +5 \ cdot y' (t) +2 \ cdot y (t)=1- e ^ {- t} \\ Initial \ conditions \ t=0; \ Rightarrow y (0)=0; y '(0)=0. \ \ End {gathered} \ right. $


Solution. $\Rightarrow$Запишем однородное ОДУ: $2 \cdot y''(t)+5 \cdot y'(t)+2 \cdot y(t) =0$
Характеристическое уравнение имеет вид: $2 \cdot \lambda ^2+5 \cdot \lambda+2=0$; Решая, имеем: $\lambda_1=-2; \ \ \lambda=0.5,$тогда:

$y_{соб}=С_1 \cdot e^{-2 \cdot t}+С_2 \cdot e^{-0.5 \cdot t},$


где $С_1, С_2$— неизвестные (пока) постоянные интегрирования.

По виду временной функции в правой части запишем $y_{вын}(t)$как:


$у_{вын}(t) =A+B \cdot e^{-t} \Rightarrow у_{вын}'(t)=-B\cdot e^{-t} \Rightarrow у_{вын}''(t)=B\cdot e^{-t} $


Подставляя в исходное уравнение, имеем:


$2\cdot B \cdot e^{-t} - 5 \cdot B \cdot e^{-t}+2\cdot A +2 \cdot B \cdot e^{-t} =1 - e^{-t} \Rightarrow \left\{ \begin{gathered} 2 \cdot A =1\\ -B=-1\ \end{gathered} \right. \Rightarrow\\ \Rightarrow \left\{ \begin{gathered} A=\frac{1}{2}\\ B=1\ \end{gathered} \right. \Rightarrow y_{вын.}(t)=\frac{1}{2} -e^{-t};$


Суммируя $y_{соб}, y_{вын}$, имеем: $y(е)=С_1 \cdot e^{-2 \cdot t}+С_2 \cdot e^{-0.5 \cdot t}+\frac{1}{2}+ e^{-t}.$


Используя 1-е начальное условие (при t=0), получаем: $0 =C_1+C_2+0.5+1$, а из 2-го начального условия имеем: $0=-2 \cdot C_1 - 0.5 \cdot C_2 -1.$


Решая систему уравнений относительно $С_1$и $С_2$, имеем: $С_1=-1/6; \ \ \ C_2=-4/3. $
Тогда окончательно:


$y(t)=- \frac{1}{6} e^{-2t}- \frac{4}{3} e^{-0.5t}+\frac{1}{2}+e^{-t};$


Что бы проверить результ, выполним моделирование процесса в SimInTech, для этого преобразуем исходное уравнение к виду:


$2 \cdot y''(t)+5 \cdot y'(t)+2 \cdot y(t) =1- e^{-t} \Rightarrow \\ \Rightarrow y''(t)=0.5 - 0.5\cdot e^{-t} - 2.5 \cdot y'(t)- y(t)$


Создадим модель SimInTech, содержащую исходное динамическое уравнение и полученное аналитическое решение, и выведем результаты на один график (см. рис. 2.3.1).


ITKarma picture
Fig. 2.3.1 - block diagram for verifying the solution


In fig. 2.3.2 the solution according to the above relation and the numerical solution of the problem in the SimInTech environment are presented (the solutions coincide and the graph lines are “superimposed” on each other).


ITKarma picture

Fig. 2.3.2 - Solution of the equation of dynamics

Related Links:


  1. Wikipedia about the Taylor series
  2. Differential equations on Match24.ru
  3. An example of creating a model of a load on a spring.
  4. Lecture start here: "Introduction to the theory of automatic control. The basic concepts of the theory of control of technical systems "
.

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